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)

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)

Answer Details

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

volume =  500

cm3
$𝑐$ $𝑐$ ${𝑚}^{}$ $3^{}$ = 0.5dm3, c = 0.005moldm−3
$𝑚$ $𝑚$ $𝑜$ $𝑙$ $𝑑$ ${𝑚}^{}$ ${-}^{}$ $3^{}$massmolarmass
$\frac{𝑚}{}$ $\frac{𝑚}{}$ $\frac{𝑎}{}$ $\frac{𝑠}{}$ $\frac{𝑠}{}$ $\frac{𝑚}{}$ $\frac{𝑜}{}$ $\frac{𝑙}{}$ $\frac{𝑎}{}$ $\frac{𝑟}{}$ $\frac{𝑚}{}$ $\frac{𝑎}{}$ $\frac{𝑠}{}$ $\frac{𝑠}{}$ =  0.0025 = mass98
$\frac{𝑚}{}$ $\frac{𝑚}{}$ $\frac{𝑎}{}$ $\frac{𝑠}{}$ $\frac{𝑠}{}$ $\frac{98}{}$ (where 98g/mol is the molar mass of H2SO4)

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

but n = 

mass of H2SO4 = 0.0025 x 98 = 0.245g.