How many moles of KF are contained in 347 g of water in a 0.175 m KF solution?
A) 1.65 × 10-2 mol KF
B) 5.04 × 10-2 mol KF
C) 6.07 × 10-2 mol KF
D) 3.22 × 10-2 mol KF
E) 1.98 × 10-2 mol KF
No. of moles = molarity x weight in 1000 mL
= 0.175 X (347/1000)
= 0.175 x 0.347
= 0.0607
= 6.07 X 10-2 mol KF
Answer is C
How many moles of KF are contained in 347 g of water in a 0.175 m...
Question #12
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