IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies.
Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..
Q1.The decomposition of a certain mass of CaCO_3 gave 11.2 dm^3 of CO_2 gas as STP. The mass of KOH required to completely neutralise the gas is
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAJJ2LHvUuYh7c2SaP5knxXAFT9GlGnj-g5mylAVcfKSaRBBaqOfX6Z2TK5r2d_c-st9WiDtHFyxaKWDj1yUrCTc-kllcrCJK52gKEXYiOB-WfYa9H412-RGAH0yV3Bb5SfZez7G-0KeRq/)
Q2.The molarity (conc. In mol L^(-1)) of H_2 SO_4 solution, which has a density 1.84 g/cc at 35℃ and contains 98% by weight is
Solution
Q3. The normality of solution obtained by mixing 10 mL of N/5 HCl and 30 mL of N/10 HClis
Solution
Q4. What would you observe if you add with shaking excess dil. NaOH solution to ZnCl_2 solution?
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgAeBN9iYvqBV6x8FUCG7TgPU8SzANukh2xJZHSpe2BsfohyFHRpu0OGh3kOSmXg2xxpn7dXTXBwrXyEo47OuniX4aQXf__8Fxyc6KtZxdzuX-CehBxNbwMThp2OXuVeGBNeIqljdzvQiEq/)
Q5. Hardness of water is measured in terms of ppm (parts per million) of CaCO_3.It is the amount (in g) of CaCO_3 present in 10^6 g H_2 O. In a sample of water, 10 L required 0.56 g of CaO to remove temporary hardness of HCO_3^- Ca(HCO_3 )_2+CaO→2CaCO_3+H_2 O Temporary hardness is
Solution
Q6. To 1 L of 1.0 M impure H_2 SO_4 sample, 1.0 M NaOH solution was added and a plot was obtained as follows
The % purity of H_2 SO_4 and the slope of curve, respectively, are:
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJwaOMw9_Ir063vX8BEQQ9ukOfLDDgW3gORDFjpcTTolozo-fiTeoMh7caIx5pfkIDxDdBUEhT8CYWHER_Nv0DsgpRdlGX2MmNosSF40F3tq7MXnDowjwqslR17CtN3vByDhHivyAgxBbo/)
Q7. The Ew of an element is 13. It forms an acidic oxide which with KOH forms a salt isomorphous with K_2 SO_4. The atomic weight of element is
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh7wmlgFbzWZvID6kxBJk-ZsSuzj0NZrUyX6NUQRa9TVyJFwtK6TGA4AZ6DAn-ovBVLiiLQXaVImNccBVJBpDN_nRsaMCE1GcKJp6A79qLHdaQaunJnsw4wfOCUl9ZU1lF6243nTBIPTOlq/)
Q8. NaHC_2 O_4 is 0.1 M when neutralized with NaOH.Hence, it is --------- when oxidized with MnO_4^-/H^+
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjkxj9S9SRDdHCm8cycuCr6OdJ6G0LgTNoac8K2cYkCs3HwxXhddf2hXa0Y3j1O0Zm5eKZCEgPLNvLvEdBToC42DJC3KsoriwNUa88gP8VG8tp8sZc_ly4WzU3cy6aVxcuzOxshvFS9Ld6N/)
Q9. 0.106 g of Na_2 CO_3 completely neutralizes 40.0 mL of H_2 SO_4.Hence, normality of H_2 SO_4solution is
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg12tXpbtadG6mHQrQ9YIimxWK6mYxSFWssoUytGOouNzEZBAj4vsKtLQbGTWlQyfTHMQ12VG9gXCcrdsDiS_MGTbc_2CaW7_EpXbTt3cnDD6-KjWxDDWIKQj5NAikAzCzh3d8gMX0s1XrA/)
Q10. In diammonium phosphate (NH_4 )_2 HPO_4, the percentage of P_2 O_5 is
Solution