The temperature will be **356 K**.

So, you know that you must use the **Clausius-Clapeyron equation**. Now, you'll find this equation written is several equivalent forms, so I'll just choose one of these forms

#ln(P_1/P_2) = (DeltaH_("vap"))/R * (1/T_2 - 1/T_1)#, where

#P_1# - the vapor pressure measured at #T_1#;

#P_2# - the vapor pressure measured at #P_2#;

#DeltaH_("vap")# - the enthalpy of vaporization;

#R# - the gas constant - expressed in *Joules per mol K*;

You have everything you need to solve for #T_2#. Since the pressure measured at this new temperature will be **5.00 times** bigger than #P_1#, you can write it as #P_2 = 5 * P_1# and use it in this form in the equation.

So, plug all in and you'll get

#ln(P_1/(5 * P_1)) = (46340"J"/"mol")/(8.31446"J"/("mol" * "K")) * (1/T_2 - 1/"323 K")#

#ln(1/5) = "5573.4" * 1/T_2 - "5573.4" * 1/"323"#

#-1.6094 = "5573.4"/T_2 - 17.2252#

#15.646 = "5573.4"/T_2 => T_2 = 5573.4/15.646 = "356.2 K"#

Rounded to three sig figs, the answer will be

#T_2 = "356 K"#