A solution is prepared by dissolving 20g NaOH in 1250 mL of a solvent of density 0.8 g/mL. Then the molality of the solution is

Question

A solution is prepared by dissolving 20g NaOH in 1250 mL of a solvent of density 0.8 g/mL. Then the molality of the solution is (A) 0.2 mol kg-1 (B) 0.08 mol kg-1 (C) 0.25 mol kg-1 (D) 0.0064 mol kg-1

Tardigrade Admin · Accepted Answer

see you have 20 gram NaOH. and molar mass of it is =40 gram. so no. of moles =20 / 40=1 / 2 mole. now volume of solvent =1250 ml and density =0.8 g / ml. so mass of solvent =0.8 × 1250=1000 gram =1 kg. so molality = mole of solute/ mass of solvent in kg. =1 / 2 / 1=1 / 2 molal =0.5 molal.

Q. A solution is prepared by dissolving $20 g N a O H$ in $1250 m L$ of a solvent of density $0.8 g / m L$ . Then the molality of the solution is

$0.2 m o l k g^{- 1}$

$0.08 m o l k g^{- 1}$

$0.25 m o l k g^{- 1}$

$0.0064 m o l k g^{- 1}$

$0.5 m o l k g^{- 1}$

Q. A solution is prepared by dissolving 20gNaOH in 1250mL of a solvent of density 0.8g/mL. Then the molality of the solution is

0.2molkg−1

0.08molkg−1

0.25molkg−1

0.0064molkg−1

0.5molkg−1