The approximate molarity of a solution in mol L -1 that contains 13.50 g of NaCl dissolved in 452 mL of water is

Question

The approximate molarity of a solution in mol L -1 that contains 13.50 g of NaCl dissolved in 452 mL of water is (A) 0.25 (B) 0.51 (C) 1.0 (D) 1.2

Tardigrade Admin · Accepted Answer

Given, Mass of NaCl =13.5 g Volume of solution =452 mL Molarity of solution (M) =( text Mass of NaCl / text Molar mass of NaCl ) × (1/ text Volume of solution in (L) ) M=(13.50/58.5) × (1000/452) mol / L ⇒ M=0.51 mol / L

Q. The approximate molarity of a solution in $m o l L^{- 1}$ that contains $13.50 g$ of $N a Cl$ dissolved in $452 m L$ of water is

0.25

0.51

1.0

1.2

Given,

Mass of $N a Cl = 13.5 g$

Volume of solution $= 452 m L$

Molarity of solution (M) $= \frac{Mass of N a Cl}{Molar mass of N a Cl} \times \frac{1}{Volume of solution in (L)}$

$M = \frac{13.50}{58.5} \times \frac{1000}{452} m o l / L \Rightarrow M = 0.51 m o l / L$

Q. The approximate molarity of a solution in molL−1 that contains 13.50g of NaCl dissolved in 452mL of water is

0.25

0.51

1.0

1.2

Given, Mass of NaCl=13.5g Volume of solution =452mL Molarity of solution (M) = Molar mass of NaCl Mass of NaCl​× Volume of solution in (L) 1​ M=58.513.50​×4521000​mol/L⇒M=0.51mol/L

Q. The approximate molarity of a solution in $m o l L^{- 1}$ that contains $13.50 g$ of $N a Cl$ dissolved in $452 m L$ of water is

Given,

Mass of $N a Cl = 13.5 g$

Volume of solution $= 452 m L$

Molarity of solution (M) $= \frac{Mass of N a Cl}{Molar mass of N a Cl} \times \frac{1}{Volume of solution in (L)}$

$M = \frac{13.50}{58.5} \times \frac{1000}{452} m o l / L \Rightarrow M = 0.51 m o l / L$