Question

A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g/mol)​

Answer 1

Answer :

• Density of the solution = $\bold{1.288\:g\:mL^{-1}}$

Step-by-step explanation :

Given that,

• 6.90 moles of KOH are present in 1 litre of the solution.
• Molar mass of KOH = $\bold{56\:g\:mol^{-1}}$

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We know,

$\bold{No.\:of\:moles\:of\:KOH=\dfrac{Mass\:in\:g}{Molar\:mass}}$

$\bold{OR,}$

$\bold{Mass\:of\:KOH=No.\:of\:moles\times{}Molar\:mass}$

∴ $\bold{Mass\:of\:KOH=6.90\times{}56\:g=386.4\:g}$

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

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Now, 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,

$\implies\bold{\dfrac{100}{30}\times{}386.4\:g=1288\:g}$

Now, we know $\bold{Density=\dfrac{Mass}{Volume}}$

∴ Applying the values in the expression, we get,

Density of the solution = $\bold{\dfrac{1288\:g}{1000\:mL}}$

$\implies\bold{1.288\:g\:mL^{-1}.}$

Answer 2

Explanation:

We know that,

$\boxed{\sf M=\dfrac{\% W/W ×10×d}{Mw_2}}$

• $\sf M = Molarity \: of \: the \: solution$
• $\sf d = density \: of \: the \: solution$
• $\sf Mw_2=Molecular \: weight \: of \: KOH$

$\sf 6.9=\dfrac{30×10×d}{56}$

$\sf d = 1.288 \: g/ml$