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4 August, 15:29

If 0.0203 g of a gas dissolves in 1.39 l of water at 1.02 atmospheres at what pressure (in atm) would you be able to dissolve 0.146 g in the same amount of water?

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  1. 4 August, 17:18
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    First, we need to get n1 (no. of moles of water) : when

    mass of water = 0.0203 g and the volume = 1.39 L

    ∴ n1 = mass / molar mass of water

    = 0.0203g / 18 g/mol

    = 0.00113 moles

    then we need to get n2 (no of moles of water) after the mass has changed:

    when the mass of water = 0.146 g

    n2 = mass / molar mass

    = 0.146g / 18 g / mol

    = 0.008 moles

    so by using the ideal gas formula and when the volume is not changed:

    So, P1/n1 = P2/n2

    when we have P1 = 1.02 atm

    and n1 = 0.00113 moles

    and n2 = 0.008 moles

    so we solve for P2 and get the pressure

    ∴P2 = P1*n2 / n1

    =1.02 atm * 0.008 moles / 0.00113 moles

    = 7.22 atm

    ∴the new pressure will be 7.22 atm
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