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{:(,"Column-I",,"Column-II"),("(A)","Ato...

`{:(,"Column-I",,"Column-II"),("(A)","Atomic mass in grams","(p)","Mole"),("(B)","Gram molar mass","(q)","Gram atom"),("(C)","Avogadro's No.","(r)",6.022xx10^(23)),("(D)","22.4 L of any gas at N.T.P.","(s)","Molecular mass in grams"):}`

Answer

Step by step text solution for {:(,"Column-I",,"Column-II"),("(A)","Atomic mass in grams","(p)","Mole"),("(B)","Gram molar mass","(q)","Gram atom"),("(C)","Avogadro's No.","(r)",6.022xx10^(23)),("(D)","22.4 L of any gas at N.T.P.","(s)","Molecular mass in grams"):} by CHEMISTRY experts to help you in doubts & scoring excellent marks in Class 11 exams.

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Here each question contains statements given in two columns which have to be matched. Statements in Column I are labelled as A,B,C and D. Whereas statements in column II are labelled as p,q,r and s. The answers to these questions have to be appropriately bublled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-p, then the correctly bubbled 4xx4 matrix should look like the following. {:(,"Column I",,"Column II"),((A),"Atomic mass in grams ",p,"Mole"),((B),"Grams molar mass",q,"Grams atoms"),((C),"Avogadro's number",r,"Molecular mass in grams"),((D),"22.4 litres of any gas at NTP ",s,6.022xx10^(23)):}

{:(,"Column I",,"Column II"),("(A)","1 Hectogram","p)",10^(-6)" gram"),("(B)","1 Decagram","q)",10^(2)" gram"),("(C)","1 milli gram","r)","10 gram"),("(D)","1 micro gram","s)",10^(-3)" gram"):}

Knowledge Check

  • {:(,"List-I",,"List-II"),("(P)","12 g of carbon",(1),"Atomic mass unit"),("(Q)","28 g of nitrogen",(2),"Gram atom"),("(R)","22400 cc",(3),"Gram molar volume"),("(S)",1.67xx10^(-24)g,(4),"Gram molecule"):}

    A
    P-2, Q-4, R-3, S-1
    B
    P-1, Q-4, R-3, S-2
    C
    P-3, Q-2, R-1, S-4
    D
    P-4, Q-1, R-2, S-3

  • one gram mole of a gas at N.T.P. occupies 22.4 L. This fact was derived from

    A
    Law of gaseous volumes
    B
    Avogadro's hypothesis
    C
    Berzerlius hypothesis
    D
    Dalton's hypothese

  • One gram mole of a gas at N.T.P. occupies 22.4 litres. This fact was derived from

    A
    Law of constant composition
    B
    Avogadro's hypothesis
    C
    Gay Lussac's law of combining volume
    D
    Dalton's atomic theory

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    Gram Atomic Mass|Gram Molecular Mass|Mole Concept

    Match the statements in Column I labeled as (a), (b), (c ), and (d) with those in Column II labeled as (p), (q), (r ), and (s). Any given statement in Column I can have correct matching with one or more statements in Column II. {:("Column I",,"Column II"),("(a) Lithium atom",,"(p) 54.4 eV"),("(b) Helium atom",,"(q) 13.6 eV"),("(c) Beryllium atom",,"(r) 122 eV"),("(d) Hydrogen atom",,"(s) 217.6 eV"):}

    State the following : (i) atomic mass (ii) gram atomic mass (iii) gram molar volume.

    What is the mass in grams of : (i) 6.022 xx 10^(23) atoms of oxygen (ii) 1.0 xx 10^(23) molecules of H_(2)S (iii) 6.022 xx 10^(23) molecules of oxygen ?

    30 g of element x contains 18.069 xx 10^(23) atoms of x. Calculate gram-molecular mass of x_(2)

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