Question

14.15 g of NaOH was dissolved in water and 300mL of water was added, What is M?
1.5 M NaCl was made when how many moles of NaCl was dissolved in 500mL of water?
3.1 M NaOH was made when what mass of NaOH was dissolved in 1L of water?
116g of NaCl was used to create a 5M solution, what was the volume?

Answer 1

To find a solution we will use such formulas

n(moles) = m(grams) / M_W(g/mol)
C(concentration in mol/L) = n(moles)/V(volume in L)
All volums in mL must be transformed to L.

a) 14.15 g of NaOH was dissolved in water and 300 mL of water was added, What is [b]M?
[/b]
The formula for concentration is

C(mol/L) = n(moles)/V(L)
n(NaOH) = m(NaOH)/M_W(NaOH) = 14.15/(23 + 16 +1) = 0.354 moles
C(NaOH) = n(NaOH)/V(water) = 0.354 / 0.300 = 1.18 M

Answer: C(NaOH) = 1.18 M

b) 1.5 M NaCl was made when how many moles of NaCl was dissolved in 500 mL of [b]water?
[/b]
From the formula for concentration

n (NaCl) = C(NaCl) x V(water) = 1.5 x 0.5= 0.75 moles

Answer: n (NaCl) = 0.75 moles

c) 3.1 M NaOH was made when what mass of NaOH was dissolved in 1 L of water?

n(NaOH) = C(NaOH)V(water) = 3.1 x 1.0 = 3.1 moles
m(NaOH) = n(NaOH)M_W(NaOH) = 3.1 x (23 + 16 +1) = 0.0775 grams ( or 77.5 mg)

Answer: m(NaOH) = 0.0775 grams

d) 116g of NaCl was used to create a 5Msolution, what was the volume?

n(NaCl) = m (NaCl) / MW (NaCl) = 116/ (23.0 + 35.5) = 1.98 moles
V(water) = n (NaOH) / C(NaOH) = 1.98 / 5 = 0.396 L

Answer: V(water) = 0.396 L