Question

What mass of solid Al2(SO4)3 is required to make 58 mL of a 0.010 M solution of Al2(SO4)3?

Answer 1

GIVEN :

$\bullet$ Volume of solution of Al₂(SO₄)₃ = 58ml

$\bullet$ Molarity of solution of Al₂(SO₄)₃ = 0.010 M

$\sf \bold{To\:Find\::}$

Mass of solid Al₂(SO₄)₃

$\sf \bold{Formula\:used\::}$

$\sf \bullet\: Molarity \:=\:\dfrac{Number \:of\:mole\:solute}{Volume\:of\: solution\: in\: L } \\ \\ \sf\bullet\: \: Number\: of\: moles \:=\: \dfrac{Mass}{Molar\:mass}$

$\sf \bf Solution\::$

First of all we will find molar mass of Al₂(SO₄)₃

➝ Molar mass of Al₂(SO₄)₃ = [ 2× atomic mass of Al ] + 3 × [ (1× atomic mass of S) + (4× atomic mass of O) ]

➝ Molar mass of Al₂(SO₄)₃ = [ 2×27] + 3×[(1×32)+(4×16)]

➝ Molar mass of Al₂(SO₄)₃ = 54 + 3×( 32+64)

➝ Molar mass of Al₂(SO₄)₃ = 54 + 3×96

➝ Molar mass of Al₂(SO₄)₃ = 54 + 288

➝ Molar mass of Al₂(SO₄)₃ = 342 g/mol

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Now we will find number of mole of Al₂(SO₄)₃ present in given solution

$\sf \: Molarity \: = \dfrac{Number\: of\: moles\: of\: Al_2(SO_4)_3}{Volume\: of \: solution\: in \: L}\\\\\\ \sf \implies \: 0.01 \: =\:\dfrac{Number\: of\: moles\: of\: Al_2(SO_4)_3}{\dfrac{58}{1000}}\\\\ \\ \sf \implies \: 0.01 \: \times \dfrac{58}{1000}\: = Number\: of\: moles\: of\: Al_2(SO_4)_3\\ \\ \\ \sf \implies \:Number\: of\: moles\: of\: Al_2(SO_4)_3 \: = \: 58 \times 10^{-5} mol$

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Mass of Al₂(SO₄)₃ = Number of mole × molar mass

$\sf \implies \: Mass\: of \:Al_2(SO_4)_3\: = \: 58 \times 10^{-5} \times 342 g\\\\ \sf \implies \: Mass\: of \:Al_2(SO_4)_3\: = \: 19836 \times 10^{-5} g\\\\\sf \implies \: Mass\: of \:Al_2(SO_4)_3\: = \bold{0.19836\: grams}$

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$\sf \bold{ANSWER\:\::\:\:\:0.19836\:grams}$