Question

0.10 mol of sodium is reacted completely with water. The resulting aqueous solution formed is reacted with 1.5mol dm^-3 sulphuric acid. What is the volume of sulphuric acid required for the neutralisation?
i) (0.10x1000)/1.5 cm^3
ii) (1.5x1000)/0.10 cm^3
iii) (0.10x1000)/(2x1.5) cm^3

Answer 1

Idk emcrazy! I'm sorry. :( I'm taking chem this year too

Answer 2

It's ok. I guess I'll have to wait. >_<

Answer 3

I'll give you a medal while you wait :)

Answer 4

Haha. Thanks! :)

Answer 5

is it X(constant) or multiplying sign

Answer 6

answer is third one...im sure about it

Answer 7

\[2Na + 2H_{2}O \rightarrow 2NaOH+H_{2}\] (when sodium reacts with water)

u get equal no of moles in Na and NaOH =in this case O.1 mole of NaOH obtained

the reaction between NaOH and H2SO4
\[2NaOH+H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + H_{2}O\]
then u get 0.05 moles of H2SO4 to react with NaOH

then apply \[n=\frac{c \times v}{1000}\]
\[\frac{0.05 \times 1000}{1.5} \rightarrow \frac{50}{1.5} \times 2 \rightarrow \frac{100}{3}\]

Answer 8

1.5 mol*dm^-3=1.5 mole/dm^3=1.5 mol/Lt is n;t that so???

Answer 9

@thushananth01 : I worked out (iii) as well but the book was telling me (ii). So I thought maybe I was wrong. But then the books can be wrong too sometimes. Thanks anyway!
@mos1635 : Yes!

Answer 10

isn't that molarity?

Answer 11

n=C*V
V=n/c
V=0.05/1.5
V=1/30 Lt convert to cm^3
V=100/3 cm^3

Answer 12

@thushananth01 is wright

Answer 13

@emcrazy14 is wright

Answer 14

book is wrong

Answer 15

haha

Answer 16

@emcrazy14 dont just depend on answers from the book, do ur question, if u hve justified ur answer then ths grt..

Answer 17

Haha. Ok. :D