How many grams of lithium chloride are in a 3.0 Molarity solution of 0.5 L?

Explanation:

LiCl2 3 M =

in sol. 1 L there is LiCl2 3 mol

if sol. 0.5 L there is LiCl2 0.5×3 = 1.5 mol

LiCl2 = 6.9 + 35.5(2)

= 6.9 + 71

= 77.9 g

63.52g of LiCl are in a 3.0 molarity solution of 0.5L

Explanation:

Moles = Molarity x Volume

Molarity = 3.0 mol/L

Volume = 0.5L

Moles = (3.0) x (0.5)

Moles = 1.5

In order to find the grams dissolved in the solution, we must first find the molar mass of Lithium Chloride (LiCl).

Molar mass of LiCl = (6.9) + (35.45) = 42.35g/mol

Now we multiply the moles by the molar mass of LiCl to get the grams of LiCL dissolved in the solution.  

1.5 * 42.35 = 63.52g of LiCl

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm