25 g of a solute of molar mass 250 g mol-1 is dissolved in 10 mL of water to obtain a solution whose density is 1.25 g(mL)-1. The molarity and molality of the solution are respectively

Question

25 g of a solute of molar mass 250 g mol-1 is dissolved in 10 mL of water to obtain a solution whose density is 1.25 g(mL)-1. The molarity and molality of the solution are respectively (A) 0.75 M and 1 m (B) 0.8 M and 1 m (C) 1 M and 0.8 m (D) 1 M and 0.75 m

Tardigrade Admin · Accepted Answer

Molarity =( text weight of solute × 1000/ text mol.wt.of solute × text volume of solution ) =(25 × 1000/250 × 100) [∵. density of water .=1 g( mL )-1] =1 M Molality =( text weight of solute × 1000/ text mol.wt.of solute × text w.of solvent ) =(25 × 1000/250 × 100)=1 m .

Q. $25 g$ of a solute of molar mass $250 g m o l^{- 1}$ is dissolved in $10 m L$ of water to obtain a solution whose density is $1.25 g (m L)^{- 1} .$ The molarity and molality of the solution are respectively

$0.75 M$ and $1 m$

$0.8 M$ and $1 m$

$1 M$ and $0.8 m$

$1 M$ and $0.75 m$

Q. 25g of a solute of molar mass 250gmol−1 is dissolved in 10mL of water to obtain a solution whose density is 1.25g(mL)−1. The molarity and molality of the solution are respectively

0.75M and 1m

0.8M and 1m

1M and 0.8m

1M and 0.75m

Molarity = mol.wt.of solute × volume of solution weight of solute ×1000​ =250×10025×1000​ [∵ density of water =1g(mL)−1] =1M Molality = mol.wt.of solute × w.of solvent weight of solute ×1000​ =250×10025×1000​=1m.

Q. $25 g$ of a solute of molar mass $250 g m o l^{- 1}$ is dissolved in $10 m L$ of water to obtain a solution whose density is $1.25 g (m L)^{- 1} .$ The molarity and molality of the solution are respectively

$0.75 M$ and $1 m$

$0.8 M$ and $1 m$

$1 M$ and $0.8 m$

$1 M$ and $0.75 m$