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What is the mass of solute in 500\(cm^{3}\) of 0.005\(moldm^{-3}\) \(H_{2}SO_{4}\)? ( H = 1,...

Chemistry
WAEC 2018

What is the mass of solute in 500\(cm^{3}\) of 0.005\(moldm^{-3}\) \(H_{2}SO_{4}\)? ( H = 1, S = 32.0, O=16.0)

  • A. 0.490g
  • B. 0.049g
  • C. 0.245g
  • D. 0.0245g
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Correct Answer: Option C
Explanation

Using 

n = cv where n = no of mole, c = molar concentration(mol/dm3) and v = volume( dm3)

volume =  500\(cm^{3}\) = 0.5dm3, c = 0.005\(moldm^{-3}\)

number of mole(n) = c x v = 0.5 X 0.005 = 0.0025mol.

but n = \(\frac{mass}{ molar mass}\) =  0.0025 = \(\frac{mass}{ 98}\) (where 98g/mol is the molar mass of H2SO4)

mass of H2SO4 = 0.0025 x 98 = 0.245g.

 


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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709