Mole Fraction Question?

Question

How many grams of NaBr were added....?

At 37.0C, the vapor pressure of 355 g of water was reduced from 6.20×10−2 atm to 5.20×10−2atm by the addition of NaBr.

This is what I did:
Del P = x Po
.062 - .052 = x (.062)
x= .001 / .062
x=0.1612

Now # of moles in water:
355/18.02 = 19.7mol

Now use mole fraction:
xH2O = mol h2o / mol h20 + mol NaBr
mol NaBr = mol h20 / xH20 (xH20)
mol Nabr = 1 / .16
mol = 6.25

Convert moles to grams
6.25 x 103 = 643.75grams

IS this right?
Thanks!

My answer isnt correct.

Answer 1

5.20 *10-2 = [ (19.7/ 19.7 +2X ] *6.20*10-2 solve this equation for x , then convert value of x to mass of NaBr.

Answer 2

seems correct