What mass of sucrose, C12H22O11, should you dissolve in 100.0 g of water to obtain a solution in which the mole fraction of C12H22O11 is 0.124?

the sucrose mass is 264.g

Explanation:

The computation of the sucrose mass is shown below;

The Molar mass of is  342.3 g/mol

And, the molar mass of is 18g/mol

The mass of be Y

Now

The moles of is

= Mass of ÷  molar mass of

= Y ÷ 342.3 mol

And,

The moles of is

= Mass of ÷  molar mass of

= 100.0 ÷ 18

= 5.56mol

Now

The molar fraction of is

= Moles of ÷  (moles of + moles of )  = 0.124

= (Y ÷ 342.3) ÷ [(Y ÷ 342.3) + 5.56] = 0.124

Y ÷ 342.3 = 0.124 × [(Y ÷ 342.3 + 5.56]

Y = 0.124 × Y + (0.124 × 5.56 × 342.3)

Y = 269.4 g

hence, the sucrose mass is 264.g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i'm wiling to try and you : )