Question

H=1.01
What is the molality of a solution containing 75.3 grams of KCI
dissolved in 100.0 grams of water?
K= 39
Cl = 35.45

Answer 1

Answer:

Molality of solution=10.11 m

Explanation:

We are given that

Given mass of KCl(WB)=75.3 g

Given mass of water (WA)=100 g=100/1000=0.1 kg

1 kg=1000 g

Molar mass of H=1.01 g

Molar mass of K=39g

Molar mass of Cl=35.45 g

We have to find the molality of a solution.

Molar mass of KCl(MB)=39+35.45

Molar mass of KCl(MB) =74.45 g

Number of moles of solute (KCl)=$\frac{given\;mass}{molar\;mass}=\frac{W_B}{M_B}$

Number of moles of solute (KCl)=$\frac{75.3}{74.45}$

Number of moles of solute (KCl)=1.011 moles

Molality of solution

=$\frac{number\;of\;moles\;of\;solute}{mass\;of\;solvent}$

Using the formula

Molality of solution=$\frac{1.011}{0.1}$

Molality of solution=10.11 m