Sodium cyclopentadienide exists as NaC5H5.(THF)x. To determine the value of x, an experiment is performed by dissolving a small amount of the sample in water and titrating the formed NaOH with standard HCl solution.NaC5H5.(THF)x+H2O→ NaOH+C5H6+x THF Molecular formula of THF is C4H8O .The following result data was collected from such an experiment: Trial 1Trial 2Weight of NaC5H5.(THF)x0.52 g0.41 gVolume of 0.1 M HCl required to titrate NaOH 10 mL 7.88 mLThe value of x is_____.[Atomic weights: H = 1, C = 12, Na = 23]

NaC5H5.(THF)x+H2O→ NaOH+C5H6+x THFnNaC5H5.(THF)x=nNaOH⇒0.5288+72x=10×0.11000⇒x=6

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Sodium cyclopentadienide exists as $N a C_{5} H_{5} . {(T H F)}_{x}$ . To determine the value of x, an experiment is performed by dissolving a small amount of the sample in water and titrating the formed NaOH with standard HCl solution.
$N a C_{5} H_{5} . {(T H F)}_{x} + H_{2} O \overset{}{\to} N a O H + C_{5} H_{6} + x T H F$
Molecular formula of THF is $C_{4} H_{8} O$ .
The following result data was collected from such an experiment:

Trial 1
Trial 2
Weight of $N a C_{5} H_{5} . {(T H F)}_{x}$
0.52 g
0.41 g
Volume of 0.1 M HCl required to titrate NaOH

10 mL

7.88 mL
The value of x is_____.
[Atomic weights: H = 1, C = 12, Na = 23]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 6.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$N a C_{5} H_{5} . {(T H F)}_{x} + H_{2} O \overset{}{\to} N a O H + C_{5} H_{6} + x T H F$

$n_{N a C_{5} H_{5} . {(T H F)}_{x}} = n_{N a O H} \Rightarrow \frac{0.52}{88 + 72 x} = \frac{10 \times 0.1}{1000} \Rightarrow x = 6$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!