Question

How many grams of solute would you use to prepare the following solutions?

1. 349.5 mL of 1.05 M NaOH 2. 1.35 L of 0.471 M glucose (C6H12O6)

Answer 1

Answer:

Step 1: Explanation:

Molarity is defined to be the number of moles of solute divided by the volume  of solution(in L )

i.e,   Molarity = Number of moles of solute / volume of solution ( in liters)

Step 2: Calculate the moles

Given,

(1) Volume = 349.5 mL = 0.3495 L [ because 1 L=1000 mL so, 349.5 mL = 349.5 mL  × ( 1 L / 1000 mL) = 0.3495 L ]

Molarity = 1.05 M = 1.05 mol / L [ because M = mol/L]

Since we know ,

Molarity = moles of solute / volume of solution ( in L)

or, moles of solute = molarity × volume of solution

moles of solute = 1.05 mol/L × 0.3495 L = 0.366975 mol

(2) Volume = 1.35 L

Molarity = 0.471 M = 0.471 mol / L [ because M = mol/L]

moles of solute = 0.471 mol/L × 1.35 L = 0.63585 mol

Step 3: Calculate the mass in grams in solute

Now convert moles to mass by using molar mass

we know, moles = mass / molar mass

so , mass = moles × molar mass

(1) Molar mass of NaOH = 39.997 g/mol

moles we got = 0.366975 mol

mass = 0.366975 mol × 39.997 g/mol = 14.6779 gram

so number of grams of solute = 14.6779 grams

(2) Molar mass of glucose = 180.156 g/mol

moles we got = 0.63585 mol

mass = 0.63585 mol × 180.156 g/mol = 114.5522 gram

so number of grams of solute = 114.5522 grams