What is solute mass and mol solute of #NaHCO_3# which has the volume solution 250.0 mL and molarity 0.100 mol/L?

1 Answer
Jarni Renz
Jan 20, 2018

#a. eta=0.025mol#
#b. m=2.10g#

Explanation:

Given the molarity #(M)# and the volume #(V)# of the solution, the number of moles solute#(eta)# can be calculated through the formula shown below. Rearrange the formula to isolate the unknown variable, the #eta#;

#M=eta/V#
#eta=MxxV#

where:

#M=(0.100mol)/(L)#; the concentration
#V=250.0ml#

#eta=(0.100mol)/cancel((L))xx250.0cancel(ml)xx(1cancel(L))/(1000cancel(ml))#
#eta=0.025mol#

Now, since #eta# is already obtained, mass of the solute can also be calculated using the formula described below. Rearrange formula to isolate the unknown variable.

#eta=("mass"(m))/("Molar mass"(Mm))#
#m=etaxxMm#

where:

#eta=0.025mol#; as obtained from the preceding calculation
#Mm=(84.01g)/(mol)#; is obtainable from the periodic table.

#m=0.025cancel(mol)xx(84.01g)/cancel(mol)#
#m=2.10g#

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