# A bottle of cold drink contains 200ml liquid in which ##”CO”_2## is 0.1 molar. Suppose ##”CO”_2## behaves like an ideal gas, the volume of dissolved ##”CO”_2## at S.T.P is?

##”0.45 L”##

The idea here is that you need to use the and volume of the solution to determine how many moles of carbon dioxide you have dissolved in the solution.

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A bottle of cold drink contains 200ml liquid in which ##”CO”_2## is 0.1 molar. Suppose ##”CO”_2## behaves like an ideal gas, the volume of dissolved ##”CO”_2## at S.T.P is?

Once you know that, use the to calculate the volume that many moles would occupy under STP conditions.

So, molarity is defined as the number of moles of , which in your case is carbon dioxide, ##”CO”_2##, present in one liter of solution.

In this regard, a ##”0.1-M”## solution will contain ##0.1## moles of carbon dioxide for every liter of solution. This means that your sample will contain

##200 color(red)(cancel(color(black)(“mL”))) * (1color(red)(cancel(color(black)(“L”))))/(10^3color(red)(cancel(color(black)(“mL”)))) * “0.1 moles CO”_2/(1color(red)(cancel(color(black)(“L”)))) = “0.02 moles CO”_2##

Now, STP conditions are currently defined as a pressure of ##”100 kPa”## and a temperature of ##0^@”C”##. Under these specific conditions, one mole of any ideal gas occupies ##”22.7 L”##.

This is known as the at STP. Use this value to determine the volume occupied under STP conditions by your sample

##0.02color(red)(cancel(color(black)(“moles CO”_2))) * “22.7 L”/(1color(red)(cancel(color(black)(“mole CO”_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(“0.45 L”)color(white)(a/a)|)))##

I’ll leave the answer rounded to two , but keep in mind that you only have one sig fig for your values.

SIDE NOTE More often than not, the molar volume of a gas at STP will correspond to the old definition of STP conditions, which was a pressure of ##”1 atm”## and a temperature of ##0^@”C”##.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies ##”22.4 L”##. If these are the values given to you, simply redo the last calculation by using ##”22.4 L”## instead of ##”22.7 L”##.

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