What mass of barium chloride is required to make a #100*cm^3# volume of #0.25*mol*L^-1# solution with respect to #BaCl_2(aq)#?

Question

Avery Adams · Accepted Answer

Approx. #5*g#.........

#"Concentration"="Moles of solute (mol)"/"Volume of solution (L)"# And thus the product #"volume"xx"concentration"="moles of solute."# Thus,... #n_(BaCl_2)=0.25*mol*L^-1xx100*cm^3xx10^-3*L*cm^-3# #=0.025*mol# #BaCl_2# And since #"no. of moles"="mass"/"molar mass"# #"mass"=0.025*cancel(mol)xx208.23*g*cancel(mol^-1)=??g# All I have done here is to use the given quotient that defines #"concentration"#, and divided or multiplied to give the quantity I want. As a check on my arithmetic, I included the units of each quantity. The fact that I clearly got an answer in #"grams"# for a question that asked for an answer in #"grams"# persuades me that I got the order of operations right (for once!).

Caleb Adams · Answer

undefined To calculate the mass of barium chloride required, you can use the formula:

mass = molarity × volume × molar mass

Given:
Molarity (M) = 0.25 mol/L
Volume (V) = 100 cm^3 = 0.1 L (since 1 L = 1000 cm^3)
Molar mass of BaCl2 = 137.327 g/mol + 2(35.453 g/mol)

Calculate:
mass = 0.25 mol/L × 0.1 L × (137.327 g/mol + 2(35.453 g/mol))

Luna Chen · Answer

undefined To calculate the mass of barium chloride required, use the formula:

$$ \text{mass} = \text{molarity} \times \text{volume} \times \text{molar mass} $$

Given:
Molarity ($ M $) = 0.25 mol/L
Volume ($ V $) = 100 cm$^3$ = 0.1 L
Molar mass of BaCl$_2$ = 137.34 g/mol (for Ba) + 2 * 35.453 g/mol (for Cl) = 208.24 g/mol

$$ \text{mass} = 0.25 \, \text{mol/L} \times 0.1 \, \text{L} \times 208.24 \, \text{g/mol} $$

$$ \text{mass} = 5.206 \, \text{grams} $$

Therefore, approximately 5.206 grams of barium chloride is required to make a 100 cm$^3$ volume of 0.25 mol/L solution with respect to BaCl$_2$ (aq).