What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

Question

What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL? (A)  4 mol L-1 (B)  20 mol L-1 (C)  0.2 mol L-1 (D)  2 mol L-1

Tardigrade Admin · Accepted Answer

Molarity (M)= ( textNo. of moles of solute / textVolume of solution (in L) ) or, Molarity = (WB × 1000/MB text Volume of solution (in mL)) =(5.85 ×1000/58.5×500) (∴ Molar mass of NaCl = 58.5 g mol-1) = 0.2 mol L-1

Q. What will be the molarity of a solution, which contains $5.85 g$ of $N a C l_{(} s) p er 500 m L$ ?

$4$ mol $L^{- 1}$

$20$ mol $L^{- 1}$

$0.2$ mol $L^{- 1}$

$2$ mol $L^{- 1}$

Molarity $(M) = \frac{No. of moles of solute}{Volume of solution (in L )}$
or, Molarity $= \frac{W _{B} \times 1000}{M _{B} Volume of solution (in m L )}$
$= \frac{5.85 \times 1000}{58.5 \times 500}$ ( $∴$ Molar mass of $N a Cl = 58.5 g m o l^{- 1}$ )
$= 0.2 m o l L^{- 1}$

Q. What will be the molarity of a solution, which contains 5.85g of NaCl(​s)per500mL?

4 mol L−1

20 mol L−1

0.2 mol L−1

2 mol L−1