Question

A solution is prepared by dissolving 10 g NaOH in 1250 mL of a solvent of density 0.8 mL/g. The molality of the solution in mol kg⁻¹ is
(a) 0.25
(b) 0.2
(c) 0.008
(d) 0.0064

Answer 1

A)

Answer: 2.05 M

Density of the solution 1.15gmL−1=MV=1000+120V→V=11201.15=973.91mL

⇒ Molarity =WM×1000VmL=120×1000×1.2560×973.91=2.05M

Hope that helps

Mark as BRAINLIEST ❤️❤️

and do FOLLOW me ✌️✌️

Answer 2

The molality of the solution is 0.25 mol/kg

Explanation:

To calculate the mass of solvent, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solvent = 0.8 g/mL

Volume of solvent = 1250 mL

Putting values in above equation, we get:

$0.8g/mL=\frac{\text{Mass of solvent}}{1250mL}\\\\\text{Mass of solvent}=(0.8g/mL\times 1250mL)=1000g$

To calculate the molality of NaOH, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (NaOH) = 10 g

$M_{solute}$ = Molar mass of solute (NaOH) = 40 g/mol

$W_{solvent}$ = Mass of solvent = 1000 g

Putting values in above equation, we get:

$\text{Molality of NaOH}=\frac{10\times 1000}{40\times 1000}=0.25mol/kg$

Learn more about molality of solution:

https://brainly.com/question/14618061

https://brainly.com/question/14557698