What is the number of moles of Na\(_H\)CO\(_3\) in 250cm\(^3\) of its solution if the solution contains 8.4 g of Na\(_H\)CO\(_3\) in 1 dm\(^3\) of the solution. [Na = 23, 0 = 16, C = 12, H = 1)
no of mole in 250\(cm^3\) of its solution is about molar concentration.
using n = cv
c = \(\frac{n}{v}\)
where c = molar concentration, n = number of moles and v = volume in \(dm^3\)
number of mole of Na\(_H\)CO\(_3\) = mass of Na\(_H\)CO\(_3\) / molar mass of Na\(_H\)CO\(_3\) = \(\frac{8.4}{84}\) = 0.1moles
but v = 250\(cm^3\) = 250/1000 = 0.25\(dm^3\)
therefore c = 0.25 x 0.1 = 0.025mol\(dm^3\)
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