All categories

2 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

1 answer:

5 0

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

You might be interested in

Determine the relative formula mass of hexasodium difluoride using the periodic table below. A. 138 g/mol B. 176 g/mol C. 20 g/m

laila [671]

Answer:

Option B. 176g/mol

Explanation:

We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:

Hexasodium means 6 sodium atom

Difluoride means 2 fluorine atom.

Therefore, the formula for hexasodium difluoride is Na6F2.

The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.

Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:

Molar mass of Na = 23g/mol

Molar mass of F = 19g/mol

Relative formula mass Na6F2 = (23x6) + (19x2)

= 138 + 38

= 176g/mol

Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol

3 0

2 years ago

Read 2 more answers

Treating (CH3)3C-Cl with a mixture of H2O and CH3OH at room temperature would yield: A) CH2=C(CH3)2 B) (CH3)3COH C) (CH3)3COCH3

suter [353]

Answer:

All of the choices.

Explanation:

As the reaction is involving with a mixture of H₂O and CH₃OH, these two reagents can work as nucleophyle of the reaction. Both of them, are polar and promoves a Sn1/E1 reaction. When it reacts with water it will produce product B); it will form product C) when it reacts with methanol, and product A) will be formed when the reaction undergoes an E1 reaction.

In this case, the only way to show you this, it's doing the mechanism of reaction for each product. Picture attached show the mechanism for the formation of all these products.

5 0

2 years ago

Glycerol (C3H8O3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. Consider that you have an aqueous solution that contain

pychu [463]

Answer:

The vapor pressure of the solution at 25°C is 26.01 Torr

Explanation:

This is a usual excersise of colligative properties. In this case we apply the vapor pressure lowering formula:

ΔP = Xst . P°

Where ΔP is the diferrence between Pressure of solution - Pressure of pure solvent.

And Xst the molar fraction.

P° is Pressure of pure solvent.

So the formula will be:

Pressure Solution - P° = Xst . P°

Pressure Solution - 23.8 Tor = Xst . 23.8 Torr

Xst : Mole fraction ( Moles of solute or solvent / Total moles)

34.4 % m/m means that in 100 g of solution I have 34.4 g of solute

If I have 34.4 g of solute and the mass of 100 g in solution, I can know the mass of solvent, and finally the moles.

100 g solution - 34.4 g solute = 65.6 g (mass of solvent)

Molar mass of water : 18 g/m

Moles of water: Mass of water / Molar mass

65.6 g / 18g/m = 3.64 moles

Moles of glycerol : Mass glycerol / Molar mass glycerol

34.4 g / 92.1 g/m = 0.373 moles

Total moles: moles of glycerol + moles of water

0.373 m + 3.64 m = 4.01 m

So Xst = 0.373 m / 4.01 m → 0.093

Pressure Solution - 23.8 Tor = 0.093 . 23.8 Torr

Xst HAVE NO UNITS

Pressure Solution = (0.093 . 23.8 Torr ) + 23.8 Tor

Pressure Solution = 26.01 Torr

6 0

2 years ago

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.

Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

= 0.093155 M

To 4 significant figure = 0.0932 M

3 0

2 years ago

Read 2 more answers

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

2 years ago