The experimental apparatus represented above is used to demonstrate the rates at which gases diffuse. When the cotton balls are
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Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span>
= 4.8 mole * 98 g/mole = 470g
m(H2SO4) which is =<span>470g
</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
According to that: </span>
= 4.8 mole * 98 g/mole = 470g
</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>