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aksik [14]
2 years ago
10

How many moles of methanol must be added to 4.50 kg of water to lower its freezing point to -11.0 ∘c? for each mole of solute, t

he freezing point of 1 kg of water is lowered 1.86 ∘
c.
Chemistry
1 answer:
DochEvi [55]2 years ago
3 0

<u>Given:</u>

Mass of solvent water = 4.50 kg

Freezing point of the solution = -11 C

Freezing point depression constant = 1.86 C/m

<u>To determine:</u>

Moles of methanol to be added

<u>Explanation:</u>

The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:

ΔTf = kf*m

where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution

ΔTf = 0 C - (-11.0 C) = 11.0 C

m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg

11.0 = 1.86 * moles of methanol/4.50

moles of methanol = 26.613 moles

Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.



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5 0
2 years ago
Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 669.1-g sam
Vedmedyk [2.9K]

Answer:

2.4 ×10^24 molecules of the herbicide.

Explanation:

We must first obtain the molar mass of the compound as follows;

C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1

We know that one mole of a compound contains the Avogadro's number of molecules.

Hence;

169 g of the herbicide contains 6.02×10^23 molecules

Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.

7 0
2 years ago
Given the following balanced reaction of hydrochloric acid and oxygen gas forming chlorine gas and water, how many grams of hydr
nadya68 [22]
The ratio of moles of reactants to moles of products can be seen from the coefficients in a balanced equation. In our case  4 moles of hydrochloric acid reacts with one mole of oxygen to produce two moles of chlorine and water.  So, <span> the ratio of moles of hydrochloric acid to moles of chlorine is 2:1. To determine the number moles, divide the mass by the mass of one mole. </span>
<span>Cl2 = 2 * 35.45 = 70.9 grams </span>
<span>Number of moles = 335 ÷ 70.9 </span>
<span>This is approximately 4.72 moles. The number of moles of hydrochloric acid is twice this number. </span>

<span>Mass of one mole = 1 + 35.46 = 36.45 grams </span>
<span>Total mass = 2 * (335 ÷ 70.9) * 36.45 </span>
<span>This is approximately 344.45 grams. 
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6 0
2 years ago
Read 2 more answers
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
mafiozo [28]

Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

B. 0.04 mol

C. 0.997 mol

D. 0.139 mol

E. 0.0681 mol

Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm  

= 100007.775 Pa

Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

7 0
2 years ago
6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13
vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as C_7 H_6 O_3</em>

<em></em>

<u>Explanation:</u>

Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$

mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$

So the empirical formula is C_7 H_6 O_3

Empirical formula mass

=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

=1 \times C_7 H_6 O_3

Compound X  = C_7 H_6 O_3  is the Answer

8 0
2 years ago
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