Question

Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

Answer 1

Answer:  4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$  

mass of  $Fe_2(SO_4)_3$ = $moles\times {\text {Molar Mass}}=0.0105\times 399.88g/mol=4.22g$

Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$