Calculating mass of krypton from mole fraction
Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:

$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$

If we can get a second equation in both variables, you can solved the system for each variable.

You were able to determine the mole fractions $\chi$:

$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$

$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$

Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.


  From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).


$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$

$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$
$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$
$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$


  From mole fractions


$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$
$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$


  Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$.