Sulphur

Step 1: Use the osmotic pressure formula.

The osmotic pressure equation is: $%5CPi%20%3D%20i%20M%20R%20T$ where: - $%5CPi%20%3D%200.82$ atm (osmotic pressure), - $i%20%3D%202.5$ (Van't Hoff factor for CaCl $_2$ ), - $R%20%3D%200.082$ L atm mol $%5E%7B-1%7D$ K $%5E%7B-1%7D$ , - $T%20%3D%2027%5E%5Ccirc%20C%20%3D%20273%20%2B%2027%20%3D%20300$ K, - $V%20%3D%202.5$ L.

Step 2: Calculate the molarity ( $M$ ).

Rearranging the equation: $M%20%3D%20%5Cfrac%7B%5CPi%7D%7Bi%20R%20T%7D$ Substituting the values: $M%20%3D%20%5Cfrac%7B0.82%7D%7B2.5%20%5Ctimes%200.082%20%5Ctimes%20300%7D$ $M%20%3D%20%5Cfrac%7B0.82%7D%7B61.5%7D%20%3D%200.0133%20%5C%2C%20%5Ctext%7Bmol%2FL%7D$

Step 3: Calculate the moles of CaCl $_2$ .

Since $M%20%3D%20%5Cfrac%7B%7B%5Ctext%7Bmoles%20of%20solute%7D%7D%7D%7B%7B%5Ctext%7Bvolume%20in%20liters%7D%7D%7D$ , we get: $%5Ctext%7BMoles%20of%20CaCl%7D_2%20%3D%200.0133%20%5Ctimes%202.5%20%3D%200.03325%20%5C%2C%20%5Ctext%7Bmoles%7D$

Step 4: Convert moles to grams.

Molar mass of CaCl $_2$ : $40%20%2B%202(35.5)%20%3D%20111%20%5C%2C%20%5Ctext%7Bg%2Fmol%7D$ Mass of CaCl $_2$ dissolved: $x%20%3D%200.03325%20%5Ctimes%20111%20%3D%203.69%20%5Capprox%203.7%20%5C%2C%20%5Ctext%7Bg%7D$

Thus, $x$ is 3.7 g.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Sulphur

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs