Suppose a lab procedure calls for 0.75 L of a 0.25 M CaCl2 solution. How much of a 10.0 M stock solution do we dilute? Where the diluted solution is #2 (the numerator)

V₁ = 0.019 L

Explanation:

Given data:

Molarity of stock solution = 10.0 M

Volume of stock solution = ?

Molarity of CaCl₂ = 0.25 M

Volume of CaCl₂ = 0.75 L

Solution:

Formula:

V₁ = M₂V₂ / M₁

V₁ = 0.25 M × 0.75 L / 10.0 M

V₁ = 0.1875 M.L / 10.0 M

V₁ = 0.019 L

"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"