Question

A sample of H2SO4 is 94%(W/V) and it density is 1.84g/ml.calculate the molality of solution.

Answer 1

Answer:

Explanation:here is the formula

Molarity= density*10*w/v/molar mass

Molar mass of h2so4= 98

=94*10*1.84/98=17.6

Answer 2

The molality of sulfuric acid solution is 10.66 m

Explanation:

We are given:

94% W/V of sulfuric acid

This means that 94 grams of sulfuric acid is present in 100 mL of volume

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

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

Density of solution = 1.84 g/mL

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$1.84g/mL=\frac{\text{Mass of solution}}{100.0mL}\\\\\text{Mass of solution}=(1.84g/mL\times 100.0mL)=184g$

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

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

$m_{solute}$ = Given mass of solute (sulfuric acid) = 94 g

$M_{solute}$ = Molar mass of solute (glucose) = 98 g/mol

$W_{solvent}$ = Mass of solvent (water) = [184 - 94] g = 90 g

Putting values in above equation, we get:

$\text{Molality of sulfuric acid}=\frac{94\times 1000}{98\times 90}\\\\\text{Molality of sulfuric acid}=10.66m$

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