Question

molarity of a solution containing 25.03 g of FeCl3 in 500 mL of water?

Answer 1

Answer

Here, assume the mass of the solution is 100 g.

$m = \frac{w \times 1000}{m \times v}$

$(v = \frac{w}{d} = \frac{100}{1.1} ml)$

$\: \: \: \: \: = \frac{20 \times 1000}{162 \times \frac{100}{1.1} } = 1.35m$

M = 1.35 m

hope this will help you ...

$\star \star \star \star \star$

Answer 2

The molarity of the solution is calculated by using the formula:

$Molarity(M)=\frac{Moles\ of \ the\ solute(n)}{litres\ of\ solution(V)}$

Given the mass of $FeCl_{3}$= $25.03g$

Volume(V)=$500mL=0.5L$

Number of moles of $FeCl_{3}$=$\frac{Given\ mass}{Molar\ mass}$

The molar mass of $FeCl_{3}$=$56+(35.5\times 3)=162.5$

Number of moles = $\frac{25.03}{162.5} =0.15$

Molarity=$\frac{0.15}{0.5} =0.3$

Hence, the molarity of 25.03 g of FeCl3 in 500 mL of water is 0.3M.