A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mass %%. (Use the following values: molecular weight MWsucroseMWsucrose = 342.296 g/molg/mol ; density rhosol′nrhosol′n = 1.02 g/mLg/mL ; and mass of water, mwatmwat = 948.1 gg ). Note that the mass of solute is included in the density of the solution.

1) 71.9 g/L

2) 0.221 m olal

3)  7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

"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"