1
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$15 \%$ aqueous solution of glucose (molecular weight $=180 \mathrm{~g} / \mathrm{mol}$ ) is isotonic with $8 \%$ aqueous solution containing an unknown non-dissociable solute. What is the molecular weight of the unknown solute?

A

108

B

96

C

84

D

9.6

2
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The vapour pressure of pure water is 23 mmHg . The vapour pressure of an aqueous solution, which contains 10 mass per cent of solute ' $A$ ' having molecular weight 50 is

A

0.003 atm

B

34.5 atm

C

22 atm

D

0.028 atm

3
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

What is the standard cell potential for the reaction with $K=1$ (equilibrium constant)

A

One

B

Zero

C

2.303

D

Infinity

4
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

For a first order reaction ( $A \rightarrow B$ ), the temperature ( $T$ ) dependent rate constant $(k)$ in $\mathrm{s}^{-1}$ was found to follow the equation $: \log k=\left(-\frac{20}{T}\right)+4$. The activation energy ( $E_a$ ) and pre-exponential factor ( $A$ ) respectively, are

A

$46.06 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^{-4} \mathrm{~s}^{-1}$

B

$92.12 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^4 \mathrm{~s}^{-1}$

C

$46.06 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^4 \mathrm{~s}^{-1}$

D

$9.212 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^{-4} \mathrm{~s}^{-1}$

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