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

Molality of solution=10.11 m

Step-by-step 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)=
`(given\;mass)/(molar\;mass)=(W_B)/(M_B)`

Number of moles of solute (KCl)=
`(75.3)/(74.45)`

Number of moles of solute (KCl)=1.011 moles

Molality of solution

=
`(number\;of\;moles\;of\;solute)/(mass\;of\;solvent)`

Using the formula

Molality of solution=
`(1.011)/(0.1)`

Molality of solution=10.11 m