What is the weight (in gram) of Na 2 CO 3 (molar mass =106 ) present in 250 mL of its 0.2 M solution?

Question

What is the weight (in gram) of Na 2 CO 3 (molar mass =106 ) present in 250 mL of its 0.2 M solution? (A) 0.53 (B) 5.3 (C) 1.06 (D) 10.6

Tardigrade Admin · Accepted Answer

Molarity =( text Weight of solute / text Molecular weight of solute ) × (1000/ text volume(in mL) ) 0.2=( text Weight of solute /106) × (1000/250) ∴ Weight of solute =(0.2 × 106 × 250/1000)=5.3 gm

Q. What is the weight (in gram) of $N a_{2} C O_{3}$ (molar mass $= 106$ ) present in $250 m L$ of its $0.2 M$ solution?

0.53

5.3

1.06

10.6

Molarity

$= \frac{Weight of solute}{Molecular weight of solute} \times \frac{1000}{volume(in mL)}$

$0.2 = \frac{Weight of solute}{106} \times \frac{1000}{250}$

$∴$ Weight of solute $= \frac{0.2 \times 106 \times 250}{1000} = 5.3 g m$

Q. What is the weight (in gram) of Na2​CO3​ (molar mass =106 ) present in 250mL of its 0.2M solution?

0.53

5.3

1.06

10.6

Molarity = Molecular weight of solute Weight of solute ​× volume(in mL) 1000​ 0.2=106 Weight of solute ​×2501000​ ∴ Weight of solute =10000.2×106×250​=5.3gm

Q. What is the weight (in gram) of $N a_{2} C O_{3}$ (molar mass $= 106$ ) present in $250 m L$ of its $0.2 M$ solution?

Molarity

$= \frac{Weight of solute}{Molecular weight of solute} \times \frac{1000}{volume(in mL)}$

$0.2 = \frac{Weight of solute}{106} \times \frac{1000}{250}$

$∴$ Weight of solute $= \frac{0.2 \times 106 \times 250}{1000} = 5.3 g m$