Question

A 6.90 M solution of aqueous KOH has 30% by weight of KOH . Calculate the density of solution

Answer 1

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Your Answer ----->

To calculate the density of koh solution...

A 6.90 M solution contains 6.90 mol of KOH in 1000 cm3 of the solution.

Mass of 6.90 mol of KOH = 690 H 56 = 386.4 g

A 30% solution contains 30g of KOH present in = 100 g of solution∴ 386.4 g of KOH is present in = (100 x 386.4 g)/ 30 = 1288 g of solutions

Density of KOH solution = (Mass / volume) = 1288 / 1000 =1.288 g/cm-3

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Answer 2

Hey !!

6.90 M KOH solution means 6.90 moles of KOH are present in 1 litre of tthe solution.

 But 6.90 moles of KOH = 6.90 × 56 g = 386.4 g

Thus, volume of solution containing 386.4 g KOH = 1000 mL

30% by mass of KOH solution means that 30 g of KOH are present in 100 g of the solution.

∴ Mass of solution containing 386.4 g KOH = 100/30 × 386.4 g = 1288 g

∴  Density of solution = $\frac{Mass}{Volume}$ = $\frac{1288g}{1000mL}$

        = 1.288 g mL⁻¹

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