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liraira [26]
2 years ago
5

What is the molal concentration of a solution made by dissolving 34.2 g of sucrose, c12h22o11 (molar mass 342.34 g/mol), in 125

g of water?
Chemistry
2 answers:
Anna007 [38]2 years ago
7 0
Molality is the number of moles of solute in 1 kg of solvent
number of moles of sucrose - mass of sucrose / molar mass
number of moles of sucrose - 34.2 g / 342.34 g/mol = 0.0999 mol
number of moles in 125 g of water - 0.0999 mol 
therefore number of moles in 1000 g - 0.0999 / 125 x 1000 = 0.799 mol/kg
molality of sucrose solution - 0.799 mol/kg
Alborosie2 years ago
6 0

Answer:

0.799 mol

Explanation:

Molality is the number of moles of solute in 1 kg of solvent

number of moles of sucrose - mass of sucrose / molar mass

therefore number of moles in 1000 g - 0.0999 / 125 x 1000 = 0.799 mol

Molality of sucrose solution - 0.799 mol

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avogadro's number of representative particles is equal to one___. A) kilogram B)gram C)kelvin D)mole
a_sh-v [17]

D

Avogadro's number allows us to measure the amount of atoms or molecules in one mole of a substance.

8 0
2 years ago
Read 2 more answers
The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o
dezoksy [38]

A: 6.37gNaCl

B: 1 mol NaCl

C: 58.5 g NaCl

These answers are correct on e2020 I just got them right.

6 0
2 years ago
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Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
2 years ago
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
ozzi
The number  of Ml  of  a  0.40 %w/v solution  of   ,nalorphine  that must  be injected  to  obtain  a  dose  of 1.5 mg is  calculated as  below


since M/v%   is  mass  of solute  in  grams per 100  ml

convert Mg to  g
1 g = 1000 mg  what  about  1.5 mg =?  grams
=   1.5 /1000 = 0.0015 grams


volume is therefore =  100 (  mass/ M/v%)

= 100  x(  0.0015/ 0.4) =  0.375  ML
6 0
2 years ago
An unknown metal is dropped into 127 grams of water. The temperature of the water has been raised from 25OC to 28OC. Using the s
Gnesinka [82]

Answer:

he amount of heat gained by the water is 1.59 kJ    

Explanation:

Relation between heat energy, specific heat and temperature change is as follows

Q = mCΔT

where,    Q or q = heat energy

             m = mass

             C = specific heat  =4.186J/g°C

ΔT = (28°C - 25°C) = 3°C

Now, putting the given values into the above formula as follows.

Q = mCΔT

= 127 × 4.186 × 3

= 1594.86 J or 1.59 kJ    

Therefore, we can conclude that the amount of heat gained by the water is 1.59 kJ    

5 0
2 years ago
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