Freezing point of water in Kelvin is:

273

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

293

No worries! We‘ve got your back. Try BYJU‘S free classes today!

373

No worries! We‘ve got your back. Try BYJU‘S free classes today!

393

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $273$ k
The freezing point of water in Celsius is $0^{0} C$
Now, $K e l v i n t e m p e r a t u r e = C e l c i u s t e m p e r a t u r e + 273$
$∴ K e l v i n T e m p e r a t u r e = 0 + 273$ $= 273 K$
Hence, the freezing point of water in Kelvin is $273$ k.

Suggest Corrections

Similar questions

K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1}

300^{0}

0.625 g

(C_{6} H_{12} O_{6})

102.8 g

= 273 K

K_{f}

= 1.87 K

k g

{m o l}^{- 1}

C = 12

H = 1

O = 16

- 160^{\circ} Y

- 50^{\circ} Y

20^oC corresponds to _______ .

Join BYJU'S Learning Program