Answer to Question #25348 in Inorganic Chemistry for Sarah
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?
n(moles) = m(grams) / MW(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?
The formula for concentration is
C(mol/L) = n(moles)/V(L)
n(NaOH) = m(NaOH)/MW(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?
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)MW(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
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