Question

A saline solution contains 40g of common salt and 320g of water. Calculate -
a) no. of moles of common salt present in the solution.
b) no. of molecules of water present in the solution.

Answer 1

a) Given mass of common salt (NaCl) = 40 g
Molar mass of NaCl = 58.5 g mol^-1
Number of moles (n) = $\frac{Givenmass}{Molarmass}$
n = $\frac{40}{58.5}=0.683moles$

b) Molar mass of water = 18 g mol^-1
Given mass of water = 320 g
18 g of water contains = 6.022$\times$10²³ molecules of water
320 g of water contains = $\frac{6.022\times {10}^{23}}{18}\times 320=107.05\times {10}^{23}molecules$