The correct answer is 4.

To find the mass of gas Y in the container, we can apply the ideal gas law to the mixture of gases X and Y.

1. Identify the given parameters from the image:
Molar mass of gas X:
${\stackrel{-}{M}}_X=10\text{kg/kmol}$
Molar mass of gas Y:
${\stackrel{-}{M}}_Y=20\text{kg/kmol}$
Total pressure in the container:
$P=100\text{kPa}=100\times {10}^3\text{Pa}$
Total volume of the container:
$V=10{\text{m}}^3$
Temperature of the mixture:
$T=300\text{K}$
Mass of gas X in the container:
$m_X=2\text{kg}$
Universal gas constant:
$\stackrel{-}{R}=8314{\text{J kmol}}^{-1}{\text{K}}^{-1}$

2. Calculate the number of moles of gas X:
The number of moles of gas X ($n_X$) is given by:
$n_X=\frac{m_X}{{\stackrel{-}{M}}_X}$
Substituting the given values:
$n_X=\frac{2\text{kg}}{10\text{kg/kmol}}=0.2\text{kmol}$

3. Calculate the total number of moles of the mixture:
Using the ideal gas equation for the mixture:
$PV=n_{\text{total}}\stackrel{-}{R}T$
We can solve for the total number of moles ($n_{\text{total}}$):
$n_{\text{total}}=\frac{PV}{\stackrel{-}{R}T}$
Substituting the parameters:
$n_{\text{total}}=\frac{(100\times {10}^3\text{Pa})\times \left(10{\text{m}}^3\right)}{\left(8314{\text{J kmol}}^{-1}{\text{K}}^{-1}\right)\times \left(300\text{K}\right)}$
$n_{\text{total}}=\frac{{10}^6}{2494200}\approx 0.40093\text{kmol}$

4. Find the number of moles of gas Y:
Since the total number of moles is the sum of the moles of gas X and Y:
$n_{\text{total}}=n_X+n_Y$
$n_Y=n_{\text{total}}-n_X$
$n_Y=0.40093\text{kmol}-0.2\text{kmol}=0.20093\text{kmol}$

5. Calculate the mass of gas Y:
The mass of gas Y ($m_Y$) is:
$m_Y=n_Y\times {\stackrel{-}{M}}_Y$
$m_Y=0.20093\text{kmol}\times 20\text{kg/kmol}=4.0186\text{kg}$
Rounding off to one decimal place, the mass of gas-Y in the container is 4.0 kg (which is 4 kg).